Finding Earth's Mass Using Gauss Law

In summary, the conversation discusses the possibility of finding the mass of the Earth based on its volume mass density, given by the equation p = A-Br/R. The question also brings up Gauss's law for gravitation, which states that the gravitational flux through a closed surface is proportional to the total mass within that surface. However, the asker notes that the equation and units do not seem to make sense. The conversation also provides additional information, such as the value of the gravitational constant and the volume mass density of the Earth at any distance from its center. Overall, the conversation highlights the relationship between volume mass density and gravitational flux in determining the mass of an object.
  • #1
bananasplit
9
0
Is it possible to find the mass of the Earth based on the Earth's volume mass density, roe = A-Br=R, where A = 1.42 x 104 kg/m3, B = 1.16 x 104 kg/m3, and Earth’s radius
R = 6.370 x 106 m

I know that based on Gauss Law that (closed integral) g x da = -4Gmin, where g is the total electric field due to the inside and outside of the closed surface. I don't see how this is possible.
 
Last edited:
Physics news on Phys.org
  • #2
Your equation and the units don't make sense.
 
  • #3
This was the entire question
Consider a closed surface S in a region of gravitational field g. Gauss’s law for gravitation tells us that the gravitational flux through surface S is linearly proportional to the total mass min occupying the volume contained by S. More specifically, Gauss’s law states that
(closed integral)g x da = -4Gmin :
Note that g here is the total electric field, due to mass sources both inside and outside S. The value of G, the gravitational constant, is about 6.673 x10-11 N m2/kg2.
(a) Earth’s volume mass density, at any distance r from its center, is given approximately by the function p = A-Br/R, where A = 1.42 x 104 kg/m3, B = 1.16 x 104 kg/m3, and Earth’s radius R = 6.370 x 106 m. Calculate the numerical value of Earth’s mass M. Hint: The volume of a
spherical shell, lying between radii r and r + dr, is dv = 4(pie)r2dr.
 
Last edited:

1. What is Gauss Law and how is it used to find Earth's mass?

Gauss Law is a fundamental law in electromagnetism that relates the distribution of electric charges to the resulting electric field. It can also be applied to gravitational fields, and can be used to calculate the mass of an object, such as the Earth, by observing its gravitational field and using Gauss Law to relate it to the mass.

2. What are the steps involved in using Gauss Law to find Earth's mass?

The first step is to measure the strength of the Earth's gravitational field at a known distance from its center. Then, using Gauss Law, the flux of the gravitational field through a closed surface can be calculated. This flux is directly proportional to the mass of the Earth, allowing us to solve for its mass using simple algebraic equations.

3. How accurate is the measurement of Earth's mass using Gauss Law?

The accuracy of the measurement depends on the accuracy of the initial measurements of the gravitational field strength and the distance from the center of the Earth. With precise measurements and careful calculations, the mass of the Earth can be determined with a high level of accuracy.

4. Can Gauss Law be used to find the mass of other celestial bodies?

Yes, Gauss Law can be applied to any object with a gravitational field, including other planets, stars, and galaxies. It is a powerful tool in astrophysics for determining the mass of celestial bodies that cannot be measured directly.

5. Are there any limitations to using Gauss Law to find Earth's mass?

One limitation is that it assumes a perfectly spherical and uniform distribution of mass within the Earth. In reality, the Earth's mass is not evenly distributed and there are variations in its shape. These factors can introduce some error in the calculation. Additionally, the measurement of the gravitational field at a single point may not accurately represent the entire Earth's gravitational field. Multiple measurements at different locations can help mitigate this limitation.

Similar threads

  • Introductory Physics Homework Help
Replies
26
Views
576
  • Introductory Physics Homework Help
Replies
6
Views
830
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
21
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
2K
Replies
2
Views
2K
Back
Top